求loga^b+logb^a的值已知a,b是方程(log2^x)^2-log2^x^2-2=0的两个根,求loga^b+logb^a的值.--------------我知道是通过两根之和和两根之积来做.能算出a+b和a*b的答案.请问多了log是怎么做出答案的?
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![求loga^b+logb^a的值已知a,b是方程(log2^x)^2-log2^x^2-2=0的两个根,求loga^b+logb^a的值.--------------我知道是通过两根之和和两根之积来做.能算出a+b和a*b的答案.请问多了log是怎么做出答案的?](/uploads/image/z/2805384-48-4.jpg?t=%E6%B1%82loga%5Eb%2Blogb%5Ea%E7%9A%84%E5%80%BC%E5%B7%B2%E7%9F%A5a%2Cb%E6%98%AF%E6%96%B9%E7%A8%8B%28log2%5Ex%29%5E2-log2%5Ex%5E2-2%3D0%E7%9A%84%E4%B8%A4%E4%B8%AA%E6%A0%B9%2C%E6%B1%82loga%5Eb%2Blogb%5Ea%E7%9A%84%E5%80%BC.--------------%E6%88%91%E7%9F%A5%E9%81%93%E6%98%AF%E9%80%9A%E8%BF%87%E4%B8%A4%E6%A0%B9%E4%B9%8B%E5%92%8C%E5%92%8C%E4%B8%A4%E6%A0%B9%E4%B9%8B%E7%A7%AF%E6%9D%A5%E5%81%9A.%E8%83%BD%E7%AE%97%E5%87%BAa%2Bb%E5%92%8Ca%2Ab%E7%9A%84%E7%AD%94%E6%A1%88.%E8%AF%B7%E9%97%AE%E5%A4%9A%E4%BA%86log%E6%98%AF%E6%80%8E%E4%B9%88%E5%81%9A%E5%87%BA%E7%AD%94%E6%A1%88%E7%9A%84%3F)
求loga^b+logb^a的值已知a,b是方程(log2^x)^2-log2^x^2-2=0的两个根,求loga^b+logb^a的值.--------------我知道是通过两根之和和两根之积来做.能算出a+b和a*b的答案.请问多了log是怎么做出答案的?
求loga^b+logb^a的值
已知a,b是方程(log2^x)^2-log2^x^2-2=0的两个根,求loga^b+logb^a的值.
--------------
我知道是通过两根之和和两根之积来做.
能算出a+b和a*b的答案.
请问多了log是怎么做出答案的?
求loga^b+logb^a的值已知a,b是方程(log2^x)^2-log2^x^2-2=0的两个根,求loga^b+logb^a的值.--------------我知道是通过两根之和和两根之积来做.能算出a+b和a*b的答案.请问多了log是怎么做出答案的?
由已知得:
log2^a+log2^b=2--------------1
log2^a*log2^b=-2-------------2
那么 由1得:
lga/lg2 + lgb/lg2=2
(lga+lgb)=2lg2
由2得:
(lga/lg2)*(lgb/lg2)=-2
lga *lgb=-2(lg2)^2
因此 loga^b+logb^a
= lga/lgb + lgb/lga
= [(lga)^2+(lgb)^2]/(lga *lgb)
= [(lga+lgb)^2-2lga*lgb]/(lga*lgb)
= (lga+lgb)^2/(lga*lgb) - 2
= (2lg2)^2/[-2(lg2)^2] - 2
= -2-2
=-4
应该很详细了,能看懂么